Solution 3

The left-hand side is symmetric with respect to $a,$ $b,$ $c.$ Hence, we may assume that $a \gt b gt c = 0.$ Note that replacing $(a, b, c)$ with $(a - c, b - c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to obtain the minimum possible value of the left-hand side, we may assume that $c = 0.$

with equality if and only if $\displaystyle \frac{a^2}{b^2}=\frac{b^2}{a^2},$ or equivalently, $a^4=b^4.$ Since $a,b \ge 0,$ $a = b.$ But since no two of $a,$ $b,$ $c$ are equal, $a \ne b.$ Hence, equality cannot hold. This yields

by the AM-GM inequality. Note, $x+y>y$. Thus, the last inequality is a strict one.

Acknowledgment

Leo Giugiuc has kindly communicated to me this problem, along with his two solutions. This is problem 1 from the 2017 Canada Mathematical Olympiad. The complete list of the olympiad problems has been posted at the Crux Mathematicorum facebook page. Solution 3 is the official solution; Solution 4 is by Amit Itagi.